Paper 1  Objectives  47 Questions
JAMB Exam
Year: 2007
Level: SHS
Time:
Type: Question Paper
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Managing Stress and anxiety in exams, 10 tips and strategies for Coping exam stress and test anxiety.
The goal of mocks tests is to create a benchmarking tool to help students assess their performances.
These are the three things to do in test preparation for higher grades and excellent performance.
#  Question  Ans 

1. 
If 5, 8, 6 and 2 occur with frequencies 3, 2, 4 and 1 respectively. Find the product of the modal and the median number A. 36 B. 48 C. 30 D. 40
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Detailed Solution2, 5, 5, 6, 6, ,6, 6, 8, 8 Position of median = ^{N}/_{2} = ^{8}/_{2} = 4th and 5th ∴Median = (6+6) / 2 = ^{12}/_{2} = 6 Mode the item that repeat itself most = 6 ∴Product of modal and median number = 6 x 6 = 36 

2. 
In a basket, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random. What is the probability that the fruit is either a grape or a banana A. ^{6}/_{30} B. ^{5}/_{30} C. ^{17}/_{30} D. ^{11}/_{30}
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Detailed SolutionNumber of grapes = 6Number of banana = 11 Number of oranges = 13 Total = 30 P(Grape) = ^{6}/_{30}, P(Banana) = ^{11}/_{30} P(Orange) = ^{13}/_{30} ∴P(Either grape or banana) = ^{6}/_{30} + ^{11}/_{30} = ^{17}/_{30} 

3. 
The histogram above represents the weights of students who traveled out to their school for an examination. How many people made the trip? A. 78 B. 38 C. 29 D. 69
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Detailed Solutionstudents who made the trip = 2 + 3 + 4 + 5 + 6 + 8 + 1 = 29 

4. 
A senatorial candidate had planned to visit seven cities prior to a primary election. However, he could only visit four of the cities. How many different itineraries could be considered? A. 640 B. 840 C. 520 D. 920
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Detailed SolutionNumber of itineraries = ^{7}P_{4}=\(\frac{7!}{(74)!}\\ =\frac{7!}{3!}\\ =\frac{7 \times 6 \times 5 \times 4 \times 3!}{3!}\\ =840\) 

5. 
The pie chart above illustrate the amount of private time a student spends in a week studying various subjects. Find the value of k A. 90^{o} B. 60^{o} C. 30^{o} D. 40^{o}
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Detailed SolutionK + 5K + 3K + 75 + 105 = 360 (at a point)6K + 180 = 360 6K = 360  180 6K = 180 k = 180/6 = 30^{o} 

6. 
The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old? A. ^{27}/_{40} B. ^{17}/_{20} C. ^{3}/_{30} D. ^{33}/_{40}
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Detailed SolutionP(At least 11 yrs) = P(11yrs) + P(12yrs)= 27/40 + 7/40 = 34/40 = 17/20 

7. 
What is the mean deviation of 3, 5, 8, 11, 12 and 21? A. 4.7 B. 60 C. 3.7 D. 10
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Detailed Solutionmean deviation \(= \frac{\sumx\bar{x}}{n} = \frac{28}{6}\) = 4.7 

8. 
Table: A. 15 B. 13 C. 11 D. 8
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Detailed SolutionTotal mark scored = 200∴200 = 15 + 4y  4 + 5y + 54 + 28 + 8 200 = 9y + 101 200  101 = 9y 99 = 9y ∴y = 11 

9. 
In how many ways can 6 subjects be selected from 10 subjects for an examination A. 218 B. 216 C. 215 D. 210
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Detailed Solution\(^{10}C_6 = \frac{10!}{(106)!6!}=\frac{10!}{4!6!}\\=\frac{(10\times 9\times 8\times 7 \times 6!)}{4\times 3\times 2\times 1\times 6!}\\ =210\) 

10. 
Integrate \(\frac{x^2 \sqrt{x}}{x}\) with respect to x A. \(\frac{x^2}{2}2\sqrt{x}+K\) B. \(\frac{2(x^2  x)}{3x}+K\) C. \(\frac{x^2}{2}\sqrt{x}+K\) D. \(\frac{(x^2  x)}{3x}+K\)
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Detailed Solution\(\int \frac{x^2 \sqrt{x}}{x} = \int \frac{x^2}{x}  \frac{x^{\frac{1}{2}}}{x}\\\int x  x^{\frac{1}{2}}\\ =\left(\frac{1}{2}\right)x^2  \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+K\\ =\frac{x^2}{2}2x^{\frac{1}{2}}+K\\ =\frac{x^2}{2}2\sqrt{x}+K\) 
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